Let's analyze each matrix step by step to find their determinants and determine whether they are invertible. --- ### **Matrix A** \[ A = \begin{bmatrix} 4 & 1 & 1 & 1 & 1 \\ 1 & 4 & 1 & 1 & 1 \\ 1 & 1 & 4 & 1 & 1 \\ 1 & 1 & 1 & 4 & 1 \\ 1 & 1 & 1 & 1 & 4 \\ \end{bmatrix} \] This is a **5x5 symmetric matrix** with diagonal entries 4 and off-diagonal entries 1. --- ### **Step 1: Recognize the structure** The matrix has the form: \[ A = 4I + J \] where: - \(I\) is the 5x5 identity matrix, - \(J\) is the 5x5 matrix of all ones. --- ### **Step 2: Use properties of \(J\)** - \(J\) has eigenvalues: - \(5\) (once) - \(0\) (with multiplicity 4) - The eigenvalues of \(A = 4I + J\) are: - \(4 + 5 = 9\) - \(4 + 0 = 4\) (multiplicity 4) ### **Step 3: Determine eigenvalues** Eigenvalues: - \(9\) (once) - \(4\) (four times) --- ### **Step 4: Find the determinant** The determinant is the product of eigenvalues: \[ \det(A) = 9 \times 4^4 = 9 \times 256 = 2304 \] --- ### **Step 5: Invertibility** Since \(\det(A) \neq 0\), **Matrix A is invertible**. --- ## **Summary for Matrix A:** \[ \boxed{ \det(A) = 2304 \quad \Rightarrow \text{invertible} } \] --- ## **Matrix B** \[ B = \begin{bmatrix} 4 & 1 & 0 & 0 \\ 1 & 4 & 1 & 0 \\ 0 & 1 & 4 & 1 \\ 0 & 0 & 1 & 4 \\ \end{bmatrix} \] This is a **4x4 tridiagonal matrix**. --- ### **Step 1: Recognize the pattern** The matrix has: - \(4\) on the diagonal, - \(1\) on the super- and sub-diagonals, - zeros elsewhere. --- ### **Step 2: Use recursive pattern for determinants of tridiagonal matrices** Let \(D_n\) be the determinant of the \(n \times n\) matrix of this form. For \(n=1\): \[ D_1 = 4 \] For \(n=2\): \[ D_2 = \det \begin{bmatrix} 4 & 1 \\ 1 & 4 \end{bmatrix} = 4 \times 4 - 1 \times 1 = 16 - 1 = 15 \] For \(n \geq 3\): \[ D_n = 4 D_{n-1} - D_{n-2} \] This recurrence arises because the determinant of such a tridiagonal matrix follows: \[ D_n = a D_{n-1} - b^2 D_{n-2} \] where \(a=4\), \(b=1\). --- ### **Step 3: Compute \(D_3\) and \(D_4\)** - \(D_3\): \[ D_3 = 4 D_2 - D_1 = 4 \times 15 - 4 = 60 - 4 = 56 \] - \(D_4\): \[ D_4 = 4 D_3 - D_2 = 4 \times 56 - 15 = 224 - 15 = 209 \] --- ### **Step 4: Determine the determinant** \[ \boxed{ \det(B) = D_4 = 209 } \] Since \(\det(B) \neq 0\), **Matrix B is invertible**. --- ## **Summary for Matrix B:** \[ \boxed{ \det(B) = 209 \quad \Rightarrow \text{invertible} } \] --- ## **Matrix C** \[ C = \begin{bmatrix} 1 & a & a & a \\ a & 1 & a & a \\ a & a & 1 & a \\ a & a & a & 1 \\ \end{bmatrix} \] This is a **4x4 symmetric matrix** with: - Diagonal entries: 1 - Off-diagonal entries: \(a\) --- ### **Step 1: Recognize the structure** The matrix can be written as: \[ C = (1 - a) I + a J \] where: - \(I\) is the \(4 \times 4\), - \(J\) is the \(4 \times 4\) matrix of all ones. --- ### **Step 2: Eigenvalues of \(J\)** - Eigenvalues: - \(4\) (once) - \(0\) (with multiplicity 3) ### **Step 3: Eigenvalues of \(C\)** Eigenvalues are: - For the eigenvector corresponding to eigenvalue 4 of \(J\): \[ \lambda = (1 - a) + a \times 4 = 1 - a + 4a = 1 + 3a \] - For the eigenvectors corresponding to eigenvalue 0 of \(J\): \[ \lambda = (1 - a) + a \times 0 = 1 - a \] with multiplicity 3. --- ### **Step 4: Determinant** Product of eigenvalues: \[ \det(C) = (1 + 3a) \times (1 - a)^3 \] --- ### **Step 5: Conditions for invertibility** \(\det(C) \neq 0\) when: \[ (1 + 3a) \neq 0 \quad \text{and} \quad (1 - a) \neq 0 \] which gives: \[ a \neq -\frac{1}{3} \quad \text{and} \quad a \neq 1 \] --- ## **Final conclusions:** - **Matrix A:** determinant = 2304; **invertible** - **Matrix B:** determinant = 209; **invertible** - **Matrix C:** determinant = \((1 + 3a)(1 - a)^3\) - **Invertible when:** \(a \neq -\frac{1}{3}\) and \(a \neq 1\). --- **Summary:** | Matrix | Determinant | Invertible? | |---------|--------------|--------------| | A | \(2304\) | Yes | | B | \(209\) | Yes | | C | \((1 + 3a)(1 - a)^3\) | Yes, when \(a \neq -\frac{1}{3}\) and \(a \neq 1\) | --- If you have specific values of \(a\), I can also check invertibility for those.